poj 2386（深搜或广搜）
来源：互联网 发布：招聘数据 编辑：程序博客网 时间：2024/06/06 02:29
Lake Counting
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17917 Accepted: 9069
Description
Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
Given a diagram of Farmer John's field, determine how many ponds he has.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
Output
* Line 1: The number of ponds in Farmer John's field.
Sample Input
10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.
Sample Output
3
这道题目比较基础，可以深搜也可以广搜
深搜快一点 36ms，广搜47ms。很好的练习题目。
把注释的去掉就是深搜了。
#include <stdio.h>#include <queue>using namespace std;#define N 102char map[N][N];int n,m;struct Node{ int x,y;};int dis[10][3]={ {0,1},{0,-1},{-1,0},{1,0},{-1,-1},{-1,1},{1,-1},{1,1} };void search(int a,int b){ /*map[a][b]='.'; for(int i=0;i<8;i++) { int tmpx=a+dis[i][0]; int tmpy=b+dis[i][1]; if(tmpx>0 && tmpx<=n && tmpy>0 && tmpy<=m && map[tmpx][tmpy]=='W' ){ //map[tmpx][tmpy]='.'; search(tmpx,tmpy); } }*/ Node tmp,be; be.x=a,be.y=b; queue<Node> v; v.push(be); map[a][b]='.'; while(!v.empty()) { be=v.front(); v.pop(); for(int i=0;i<8;i++) { tmp.x=be.x+dis[i][0]; tmp.y=be.y+dis[i][1]; if(tmp.x>0 && tmp.x<=n && tmp.y>0 && tmp.y<=m && map[tmp.x][tmp.y]=='W' ) { v.push(tmp); map[tmp.x][tmp.y]='.'; } } }}int main(){ int i,j,count; while(~scanf("%d %d",&n,&m)) { getchar(); count=0; for(i=1;i<=n;i++){ for(j=1;j<=m;j++){ scanf("%c",&map[i][j]);} ///缺了括号，害死人啊 getchar();} for(i=1;i<=n;i++) for(j=1;j<=m;j++) if(map[i][j]=='W') { //printf("%d %d\n",i,j); search(i,j);count++; } printf("%d\n",count); } return 0;}深搜递归写法
#include <stdio.h>#include <queue>using namespace std;#define N 102char map[N][N];int n,m;void search(int i,int j){ if(map[i][j-1]=='W') { map[i][j-1]='.'; search(i,j-1); } if(map[i][j+1]=='W') { map[i][j+1]='.'; search(i,j+1); } if(map[i-1][j]=='W') { map[i-1][j]='.'; search(i-1,j); } if(map[i+1][j]=='W') { map[i+1][j]='.'; search(i+1,j); } if(map[i-1][j-1]=='W') { map[i-1][j-1]='.'; search(i-1,j-1); } if(map[i-1][j+1]=='W') { map[i-1][j+1]='.'; search(i-1,j+1); } if(map[i+1][j-1]=='W') { map[i+1][j-1]='.'; search(i+1,j-1); } if(map[i+1][j+1]=='W') { map[i+1][j+1]='.'; search(i+1,j+1); }}int main(){ int i,j,count; while(~scanf("%d %d",&n,&m)) { getchar(); count=0; for(i=1;i<=n;i++){ for(j=1;j<=m;j++){ scanf("%c",&map[i][j]);} ///缺了括号，害死人啊 getchar();} for(i=1;i<=n;i++) for(j=1;j<=m;j++) if(map[i][j]=='W') { //printf("%d %d\n",i,j); search(i,j);count++; } printf("%d\n",count); } return 0;}
1 0
- poj 2386（深搜或广搜）
- poj 2386（深搜或广搜）
- NYOJ 27 水池数目（深搜或广搜）
- nyoj27 水池数目（深搜或广搜）
- 广搜 poj 3126
- POJ 3126 广搜
- POJ 2049 广搜
- 广搜poj.3009
- POJ 2243 广搜
- poj-2251 广搜
- poj 3414 广搜
- poj 3278 广搜
- POJ 3087 广搜
- POJ 3669 - Meteor Shower（广搜）
- POJ 1915（双向广搜）
- POJ 3984----迷宫问题（广搜）
- poj 3984 迷宫问题（广搜）
- POJ 1915 Knight Moves（广搜）
- UVALive 4128 Steam Roller（多状态最短路）
- 东软睿道实训日记——第十天
- 将图片和文字写到pdf文件中
- python学习笔记（出错）
- C#方法重载（overload）、重写（覆盖）（override）、隐藏（new）
- poj 2386（深搜或广搜）
- WIKIOI-1361 知识排名
- 领域实现更好的增长
- 死链对网站关键词排名的影响
- 【HEVC学习与研究】27、CABAC解析语法元素SAO
- 伊拉克高官儿子未赶上飞机 怒拨电话让航班返航
- Excel通用导入实现
- MyEclipse2013安装及破解教程（转载有改动，更详细一些）2014年3月亲测
- 函数指针和指针函数
